Will you get less wet running in the rain?

This is something I did one morning while waiting to take an exam that afternoon on mathematical methods and fluid mechanics. I couldn't bring myself to do any more revision but needed to keep my mind on maths. As it was raining outside I started to wonder whether it was possible to build a simple mathematical model for an object moving in the rain. What follows is a write up of my thinking that morning.

Common sense would say that one will get less wet running in the rain than walking, but this might not necessarily be true. The slower the speed travelled the more water will land on one's your head, but the faster the speed the greater the water hitting one's front. Given that, for most people, the surface area pointing vertically upwards (top of head, shoulders and chest) is much smaller than that of their surface area pointing forwards (face, front of torso and limbs), it is possible that travelling at greater speeds could in fact make one wetter. To investigate this I build a simple mathematical model for an object moving through rain and start by looking at the case when the rain is falling directly downwards (there is no wind). Later I develop the model to include the effects of wind.

Assumptions

Rain can be modelled by a constant density vector field.
Rain travels at a constant velocity downwards.
A person can be modelled by a regular cuboid.
The person moves at constant velocity through the rain.
Rain that comes in to contact with the person is assumed to be perfectly absorbed.
Other than the person there are no other surface effects that need to be considered.

The first assumption states that rather than modelling the rain as individual drops it is possible to model it spread out evenly throughout space. This appears reasonable as long as the rain droplets are significantly smaller than the object moving through them and there is no clumpiness; i.e. they are evenly spaced out. This the key assumption of the model, without this the following maths is invalid.

The second assumption asserts that rain moves at a constant speed towards the floor and doesn't change direction. While in the real world the wind will affect the direction and speed of rain for the sake of simplicity this model will ignore these effects. This assumption will make the model far simpler than including wind direction, but is an unrealistic assumption, any further models should include wind effect.

The third assumption appears to be fine for out basic initial model, further investigation is needed to confirm this for more advanced models.

The fourth assumption states that we won't be taking the acceleration of the object into account. As the velocities involved are quite small I think this is fair. Again, a more developed model may want to take this into account.

The fifth and sixth assumptions are similar; none of the water hitting the object bounces off and there is no spray onto the object from other surfaces. Both of these are appear to be okay for our simplistic model, but they don't realistically represent a person running in the rain where there will be some spray effects.

The Model

Let the rain be described by the vector field

\frac{d R}{d t} = - α k

(α > 0)

where:

$R$ is the rain vector field
and $α$ is the downwards speed of the rain

The body, $O$ , moving through the rain is a regular cuboid with corners located at $(0, 0, 0)$ , $(x, 0, 0)$ , $(0, y, 0)$ , $(x, y, 0)$ , $(0, 0, z)$ , $(x, 0, z)$ , $(0, y, z)$ and $(x, y, z)$ .

The equation for the velocity of $O$ through the rain is

\frac{d s}{d t} = β i

(β > 0)

where:

$s$ is the displacement vector
and $β$ is the speed

Given the above mathematical definitions it is possible to calculate the flux of the vector field $R$ for each face of $O$ as it moves at velocity $\frac{d s}{d t}$ by calculating the surface integrals, giving the volume flow rate through each surface. In other words how wet each surface gets per unit of time. The model has been created so only two surfaces need to be considered; the top and the front faces. That is surface $S_{top}$ with corners located at $(0, 0, z)$ , $(x, 0, z)$ , $(0, y, z)$ and $(x, y, z)$ , and the surface $S_{front}$ with corners located at $(x, 0, 0)$ , $(x, y, 0)$ , $(x, 0, z)$ and $(x, y, z)$ .

The surface integral for a face $S$ is defined as

\int_{S} F \cdot n d A

where:

$F$ is the vector field
$n$ is the unit normal vector for the surface
and $A$ is the area of the surface

For our model it is necessary to calculate how wet each face gets through time. The equation for this is

\frac{d w_{S}}{d t} = \int_{S} F \cdot n d A

where:

$w_{S}$ is the wetness of surface $S$ .

As the body is moving through a moving field the two vector equations describing velocity need to be combined, hence

F = \frac{d s}{d t} + \frac{d R}{d t}

= β i - α k

For $S_{top}$ the face normal points directly up, so

n = k

therefore

\frac{d w_{top}}{d t} = \int_{S_{top}} (β i - α k) \cdot k d A

= \int_{x}^{0} \int_{0}^{y} (β i - α k) \cdot k d y d x

= \int_{x}^{0} \int_{0}^{y} - α d y d x

= \int_{x}^{0} - α y d x

= α x y

and

w_{top} = \int α x y d t

= α x y t + C

as $w_{top} = 0$ when $t = 0$ hence $C = 0$ , the surface is dry at the start,

therefore

w_{top} = α x y t

For $S_{front}$ the face normal points directly outwards, so

n = i

therefore

\frac{d w_{front}}{d t} = \int_{S_{front}} (β i - α k) \cdot i d A

= \int_{0}^{y} \int_{0}^{z} (β i - α k) \cdot i d z d y

= \int_{0}^{y} \int_{0}^{z} β d z d y

= \int_{0}^{y} β z d y

= β y z

and

w_{front} = \int β y z d t

= β y z t + D

as $w_{front} = 0$ when $t = 0$ hence $D = 0$ , again, the surface is dry at the start,

therefore

w_{front} = β y z t

The total wetness of the body if found by summing the wetness of all the faces.

w = \sum w_{i}

= w_{top} + w_{front}

= α x y t + β y z t

= (α x + β z) y t

(1)

The ratio of rain landing on the top surface to rain hitting the front surface is

W_{ratio} = \frac{w_{top}}{w_{front}}

= \frac{α x y t}{β y z t}

= \frac{α x}{β z}

(2)

Initial Analysis

The model has produced two equations that will help to shed light on the initial question of how the speed travelled in the rain effects how wet a body gets. These are equation (1), giving the total wetness of the body at a specified time, and equation (2), giving a ratio for the proportion of rain landing on the top compared to the front of the body. I'll start with an analysis of this surface wetness ratio.

Ratio of Surface Wetness, $W_{ratio}$

Equation (2) gives a ratio for the proportion of rain landing on the top compared to that hitting the front, and depends on 4 independent variables: the speed of the rain, $\alpha$; the speed of the body, $\beta$; the thickness of the body, $x$; and the height of the body, $z$. This ratio shows that, according to the model, the two factors affecting how wet the top of the body gets, compared to the front, are the speed of the rain and the thickness of the body. Likewise, the two factors affecting the front compared to the top are the height and the speed of the body. By trying to find standard values for the rain speed, and body width and height it is possible to see how this ratio changes according to the speed the body travels at.

According to [1] the speed of falling rain depends on the size of the raindrop. Drizzle falls at 2 m/s, a 2 mm raindrop falls at 6.5 m/s, while a 5 mm raindrop falls at 9 m/s. To compare the effect of different rain speed I'll take a low speed of 3 m/s, a mid speed of 6 m/s, and a high speed of 9 m/s.
The mean height of an adult in England in 2008 was 1.68 m [2]. To compare the effects of height I'll take a short person to be 0.3 m lower than this and a tall person 0.3 m higher. Given that people aren't cuboid taking 80% of these values should give a fair value for the height components. This gives the heights 1.10 m, 1.34 m, and 1.58 m for a short, average and tall body, respectively.
To estimate people's thickness, finding statistics has been hard. There is data for waist and chest circumferences, but turning these into a reliable figure for thickness is nigh on impossible. The figures I've chosen are arbitrary but hopefully represent a reasonable estimate for the thickness of an individual. I will take the thickness for a slim, average and large person as 0.2 m, 0.35 m and 0.5 m, respectively.
The average walking speed is 3 miles per hour [3], or about 1.3 m/s. An Olympic sprinter can run 100m in under 10 s [4], an average speed of under 10 m/s.

References

[Weather Almanac for April 2003 - The Energy of a Rainshower]{#ref1}
[NHS Health Survey for England 2008]{#ref2}
[Wikipedia article on walking]{#ref3}
[IAAF list of fastest 100 m runners]{#ref4}